UVa 1103 - Ancient Messages [进制转换+DFS]

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提供三组测试数据吧：

5 3
fff
f0f
fff
f0f
fff
5 5
0fff0
0f0f0
fffff
00f00
00f00
5 5
0f0f0
0f0f0
fffff
00f00
00f00
0 0

AC输出是：

Case 1: K
Case 2: A
Case 3: W

做题的时候，发现使用ios::sync_with_stdio(false);就会WA，注释掉就AC，不知道是什么情况，正在研究中。

#include <bits/stdc++.h>using namespace std;const string dict[16] = {"0000", "0001", "0010", "0011","0100", "0101", "0110", "0111","1000", "1001", "1010", "1011","1100", "1101", "1110", "1111",};const int dir[4][2] = { { -1, 0 }, { 0, -1 }, { 1, 0 }, { 0, 1 } };const char alp[8] = { 'W', 'A', 'K', 'J', 'S', 'D' };map<char, int> cnt;char tab[256][256];int H, W, kase, cur;bool isin(const int r, const int c){return r >= 0 && r <= H + 1 && c >= 0 && c <= W + 1;}void DFS1(const int r, const int c){if (!isin(r, c) || tab[r][c]!='0') return;tab[r][c] = '-';for (int i = 0; i < 4; ++i)DFS1(r + dir[i][0], c + dir[i][1]);}void DFS2(const int r, const int c){if (!isin(r, c) || tab[r][c] != '1') return;tab[r][c] = '-';for (int i = 0; i < 4; ++i){int r1 = r + dir[i][0], c1 = c + dir[i][1];if (tab[r1][c1] == '0')++cur, DFS1(r1, c1);else DFS2(r1, c1);}}int main(){//ios::sync_with_stdio(false);while (memset(tab, '0', sizeof(tab)), cnt.clear(), cin >> H >> W, H || W){W *= 4;for (int i = 1; i <= H; ++i){string line, res; cin >> line;for (auto i : line) res += dict[i >= 'a' ? (i - 'a' + 10) : (i - '0')];memcpy(tab[i]+1, res.c_str(), W);}DFS1(0, 0);for (int i = 1; i <= H; ++i){for (int j = 1; j <= W; ++j){if (tab[i][j] != '1') continue;cur = 0;DFS2(i, j);cnt[alp[cur]]++;}}printf("Case %d: ", ++kase);for (auto i : cnt) while (i.second--) cout << i.first;cout << endl;}return 0;}

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